Problem: Simplify and expand the following expression: $ \dfrac{3}{2x - 16}- \dfrac{5}{3x - 12}- \dfrac{1}{x^2 - 12x + 32} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{3}{2x - 16} = \dfrac{3}{2(x - 8)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{5}{3x - 12} = \dfrac{5}{3(x - 4)}$ We can factor the quadratic in the third term: $ \dfrac{1}{x^2 - 12x + 32} = \dfrac{1}{(x - 8)(x - 4)}$ Now we have: $ \dfrac{3}{2(x - 8)}- \dfrac{5}{3(x - 4)}- \dfrac{1}{(x - 8)(x - 4)} $ The least common multiple of the denominators is: $ 6(x - 8)(x - 4)$ In order to get the first term over $6(x - 8)(x - 4)$ , multiply by $\dfrac{3(x - 4)}{3(x - 4)}$ $ \dfrac{3}{2(x - 8)} \times \dfrac{3(x - 4)}{3(x - 4)} = \dfrac{9(x - 4)}{6(x - 8)(x - 4)} $ In order to get the second term over $6(x - 8)(x - 4)$ , multiply by $\dfrac{2(x - 8)}{2(x - 8)}$ $ \dfrac{5}{3(x - 4)} \times \dfrac{2(x - 8)}{2(x - 8)} = \dfrac{10(x - 8)}{6(x - 8)(x - 4)} $ In order to get the third term over $6(x - 8)(x - 4)$ , multiply by $\dfrac{6}{6}$ $ \dfrac{1}{(x - 8)(x - 4)} \times \dfrac{6}{6} = \dfrac{6}{6(x - 8)(x - 4)} $ Now we have: $ \dfrac{9(x - 4)}{6(x - 8)(x - 4)} - \dfrac{10(x - 8)}{6(x - 8)(x - 4)} - \dfrac{6}{6(x - 8)(x - 4)} $ $ = \dfrac{ 9(x - 4) - 10(x - 8) - 6} {6(x - 8)(x - 4)} $ Expand: $ = \dfrac{9x - 36 - 10x + 80 - 6}{6x^2 - 72x + 192} $ $ = \dfrac{-x + 38}{6x^2 - 72x + 192}$